php - 根据下拉列表中选定的值填充 '<input>' 值

Question

我正在尝试显示 <input>值基于使用 jquery ajax 的下拉列表的选定值。

这是我的表单的标记 -

<div class="form-group">
    <label for="" class="control-label">District</label>
    <div class="element">
        <select class="form-control" id="district">
            <?php echo $option; ?>
        </select>                               
    </div>
</div>  

<div class="form-group">
    <label for="" class="control-label">Province</label>
    <div class="element">
        <input type="text" class="form-control province" placeholder="Province" value="" disabled>
    </div>
</div>

只是我需要在从上面的下拉列表中选择一个地区时显示省份输入的值。

这是我的 ajax 代码 -

$("#district").change(function(event) {
    var id = $("#district").val(); 
    //alert(data);

    /* Send the data using post and put the results in a div */
    $.ajax({
            url: "./includes/process_province.php",
            type: "post",
            data: id,
            success: function(){
                alert('Submitted successfully');
            },
            error:function(){
                alert("failure");
            }
    });

    // Prevent default posting of form
    event.preventDefault();
});

当我从下拉列表中选择一个地区时，它会出现此警报 alert('Submitted successfully'); 。但我不知道如何在我的文本字段中显示 PHP 处理值。

这是我的PHP代码来自proccess_province.php

// Require the configuration file before any PHP code:
require('./configuration.inc.php');

// Need the database connection:
require('../'.MYDB);

if(isset($_POST['id'])) {

    $province_id = (int)$_POST['id'];

        // Select exsiting data for an user and display them in the form for modify 
    $prep_stmt = " SELECT province
                                 FROM province 
                                 WHERE province_id = ?";

    $stmt = $mysqli->prepare($prep_stmt);

    if ($stmt) {
        // Bind "$userId" to parameter.
        $stmt->bind_param('i', $province_id);  
        // Execute the prepared query.
        $stmt->execute();    
        $stmt->store_result();

        // get variables from result.
        $stmt->bind_result($province);
        // Fetch all the records:
        $stmt->fetch();

        $db_province     = filter_var($province, FILTER_SANITIZE_STRING);
        //echo $province;

    } else {
        echo 'Database error';
    }   

    echo $db_province;
}

希望有人能帮助我。谢谢。

Answer 1

由于您在 ajax 文件中回显了省份的值，因此您需要根据该响应设置该值。试试这个:

     $.ajax({
            url: "./includes/process_province.php",
            type: "post",
            data: id,
            success: function(data){
                $('#province').val(data);
                alert('Submitted successfully');
            },
            error:function(){
                alert("failure");
            }
    });

还有你的标记:

<div class="form-group">
    <label for="" class="control-label">Province</label>
    <div class="element">
        <input id="province" type="text" class="form-control province" placeholder="Province" value="">
    </div>
</div>